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So, How does a dew shield... shield dew?

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#1 desertlens

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Posted 20 February 2020 - 03:08 AM

I'm thinking in terms of passive dew management. I understand what's going on with dew heaters but could use a little clarity on how the shield effects the objective and/or the air around it. Thanks.


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#2 happylimpet

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Posted 20 February 2020 - 03:16 AM

Its to do with how everything radiates heat into, and receives heat from, its environment. Normally, say in a room, everything is in balance radiating infrared and receiving the same from its surroundings. So they share the same temperature.

 

Problem is that the night sky doesnt provide this radiated heat - its effective temp is about that of the microwave background from the big bang (i understand that stars etc dont add much) so about 3K or -270c! (not sure in F sorry).

 

So a surface which sees a lot of sky will radiate heat normally but wont receive anything in return. This could be your car windscreen getting icy on a clear night, or the top surface of your scope, or lens etc. They radiate so much heat more than they receive that they drop a few degrees relative to ambient air temp, which is enough to drop below the dew point, so water condenses from the air.

 

So the way to counteract this is to place 'stuff' around the object you want to keep dew free. Even a dew shield at freezing temps still radiates a lot of infrared and will keep your scope lens (or whatever) above the dew point. Even a freezing dew shield is MUCH hotter than deep space!


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#3 orionic

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Posted 20 February 2020 - 03:17 AM

To put it one way, "A dew shield does shield as much dew as a dew shield does shield, if a dew shield does shield dew".

 

But more to the point, I understand you're just reducing the total exposure to the open sky to reduce radiative heat loss.


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#4 Redbetter

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Posted 20 February 2020 - 04:13 AM

The above descriptions have the gist of it.  Dew is forming because the surface is subcooling relative to the air.  This subcooling is caused by infrared heat radiating to the night sky (space...or at least the effective average temperature of the upper atmosphere which is quite cold, but considerably warmer than the cosmic background temperature.) 

 

Because it is radiant heat transfer, the losses can be reduced by minimizing how much sky the objective is exposed to.  A dew shield reduces the effective cross section of sky by at least an order of magnitude from what I can tell.  The dew shield itself will still subcool and there will be some radiant transfer to it as well, but that cross section will be far warmer than the upper atmosphere.


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#5 luxo II

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Posted 20 February 2020 - 06:56 AM

By slowing the loss of heat to the night sky your objective stays above the dew point, longer. Ideally long enough that it won't dew over before you're ready to pack it in.

 

An insulated dew shield is the trick, which extends a fair way back down the OTA. 


Edited by luxo II, 20 February 2020 - 06:58 AM.

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#6 John Tucker

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Posted 20 February 2020 - 07:00 AM

I disagree with most of the above, and suggest an experiment to differentiate the possibilities.

 

A simple, non-heated dew shield cannot prevent cooling of the objective lens over any meaningful period of time (1 h+) for two reasons

 

1. Radiative energy loss is fairly rapid.  If you wrap a dew shield around the objective, it simply radiates energy to the shield (which is radiating less energy back, as it is colder), and then the shield radiates it to the environment

 

2. Much if not most of the temperature drop is conductive.  The rest of your OTA is radiating energy, and so drops in temperature.  Its mass, and thus its heat capacity, is much greater than that of the objective.  The objective loses heat to the cold parts of the scope by conductive heat loss.  The tube than loses the energy gained by radiative energy loss. This is quite efficient especially if the tube is metal.

 

What actually happens is that for a limited period of time, the shield is able to reduce the amount of humid air passing over the cold surface of the lens.  The water of the condensate has to come from somewhere!

 

Thus heated dew shields are needed in the long run when air temperatures fall to near or below the dew point.

 

The experiment: There are adhesive films that attach to the sides of aquarium tanks that change color with temperature.  Take one of these and put it on the objective of an old scope, with and without a dew shield. I think you will find the tempertures are pretty much the same.

 

Similar effects are seen when you put the lid back on a plastic bin of gear out in the field, even if the lid is only placed on loosely.  10 hours later there is no way the contents of the bin are significantly warmer than the environment.  But the amount of moist air that has been in contact with your stuff is insufficent to leave everything soaking wet. 


Edited by John Tucker, 20 February 2020 - 07:10 AM.

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#7 Jon Isaacs

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Posted 20 February 2020 - 07:22 AM

A simple, non-heated dew shield cannot prevent cooling of the objective lens over any meaningful period of time (1 h+) for two reasons

 

1. Radiative energy loss is fairly rapid.  If you wrap a dew shield around the objective, it simply radiates energy to the shield (which is radiating less energy back, as it is colder), and then the shield radiates it to the environment

 

 

If I remember my physics correctly, radiative cooling is related to T4.  This is huge because the night sky is very cold in terms of radiation, something like -40C or colder.

 

The dew shield is only a few degrees cooler than the objective so there's very little radiative heat transfer between the objective and the dew shield.

 

Jon



#8 John Tucker

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Posted 20 February 2020 - 08:25 AM

If I remember my physics correctly, radiative cooling is related to T4.  This is huge because the night sky is very cold in terms of radiation, something like -40C or colder.

 

The dew shield is only a few degrees cooler than the objective so there's very little radiative heat transfer between the objective and the dew shield.

 

Jon

If it were as simple as the radiative heat transfer being equal to the temperature difference to the fourth power, wouldn't the entire OTA glow like an incandescent bulb?  Because the temperature difference between the OTA and space is larger than the temperature difference between the bulb filiment and the interior of a house.

 

Here is the way I'm thinking about it:

 

1) Once a photon leaves the surface of the objective, its gone.  The environment will not determine the rate at which photons leave.  It does determine the rate at which photons are ABSORBED by the objective

 

2) Shielded or unshielded, radiative processes lead to a net loss of heat energy by the objective, because in both cases the environment is (initially) colder than the objective, and thus less radiative

 

3) Because the metal tube has a much larger surface area than the objective, it will lose heat more rapidly and cool faster.  Because it is an excellent conductor of heat, conductive heat loss to the tube will likely be larger than direct radiative heat loss from the objective.  So items 1 and 2 are second order effects. 

 

If I wrap a 100 degree piece of heating tape around my metal tube in the vicinity of the objective, heat is transferred into the glass at a rate sufficient to keep it from dewing up, irrespective of whether or not it is shielded.  So wouldn't we expect a very rapid loss of heat in the other direction if we don't apply the heating tape and the tube cools to 40 degrees?


Edited by John Tucker, 20 February 2020 - 08:30 AM.


#9 Stellar1

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Posted 20 February 2020 - 08:33 AM

To put it one way, "A dew shield does shield as much dew as a dew shield does shield, if a dew shield does shield dew".

 

But more to the point, I understand you're just reducing the total exposure to the open sky to reduce radiative heat loss.

LMFAOOOOOOO!



#10 happylimpet

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Posted 20 February 2020 - 08:57 AM

John you make a good point that heat losses via the tube, and subsequent radiation, could be a factor. But I disagree with some other points.....

 

 

I disagree with most of the above, and suggest an experiment to differentiate the possibilities.

 

A simple, non-heated dew shield cannot prevent cooling of the objective lens over any meaningful period of time (1 h+) for two reasons

 

1. Radiative energy loss is fairly rapid.  If you wrap a dew shield around the objective, it simply radiates energy to the shield (which is radiating less energy back, as it is colder), and then the shield radiates it to the environment

 

YES IT DOES, BUT IMPORTANTLY THE OBJECTIVE ALSO RECEIVES RADIATED HEAT BACK FROM THE DEW SHIELD. REMEMBER, FREEZING TEMPS FEEL COLD TO US MAMMALS, BUT A FREEZING NIGHT IS HOT COMPARED TO MOST OF THE UNIVERSE.

 

2. Much if not most of the temperature drop is conductive.  The rest of your OTA is radiating energy, and so drops in temperature.  Its mass, and thus its heat capacity, is much greater than that of the objective.  The objective loses heat to the cold parts of the scope by conductive heat loss.  The tube than loses the energy gained by radiative energy loss. This is quite efficient especially if the tube is metal.

 

YOURE RIGHT THAT IT WILL BE AT LEAST PARTIALLY CONDUCTIVE LOSSES. YOURE ALSO RIGHT ABOUT RADIATIVE LOSSES FROM THE TUBE- SOUNDS LIKE WE AGREE!

 

What actually happens is that for a limited period of time, the shield is able to reduce the amount of humid air passing over the cold surface of the lens.  The water of the condensate has to come from somewhere!

 

BUT IF THE SURFACE IS ABOVE THE DEW POINT NO CONDENSATION WILL OCCUR. THERE IS ALWAYS SUFFICIENT CIRCULATION TO PROVIDE PLENTY OF DEW IF SURFACES ARE COLD. IN FACT AIR FLOW CAN KEEP SURFACES DEW FREE (COLD WINDY NIGHTS I NEVER HAVE ANY DEW) AS THEY KEEP SURFACES AT AMBIENT.

 

Thus heated dew shields are needed in the long run when air temperatures fall to near or below the dew point.

 

NOPE. LOOK AT OPEN AND CLOSED TUBE NEWTONIANS - THE ULTIMATE DEW SHIELD! PRIMARIES ESSENTIALLY NEVER DEW UP AND NOONE HEATS THEIR TUBES.

 

The experiment: There are adhesive films that attach to the sides of aquarium tanks that change color with temperature.  Take one of these and put it on the objective of an old scope, with and without a dew shield. I think you will find the tempertures are pretty much the same.

 

GOOD EXPT! DO IT. nOT SURE HOW MANY PEOPLE WILL STICK ADHESIVE PLASTIC ON THEIR APO LENSES THOUGH. hOWEVER THIS SHOULD SHOW THE EXPECTED RADIATIVE TEMPERATURE DROP.

 

Similar effects are seen when you put the lid back on a plastic bin of gear out in the field, even if the lid is only placed on loosely.  10 hours later there is no way the contents of the bin are significantly warmer than the environment.  But the amount of moist air that has been in contact with your stuff is insufficent to leave everything soaking wet. 

 

NOPE, THIS IS BECAUSE THE STUFF INSIDE IS KEPT WARM BY HAVING A PLASTIC COVER AT 273K ABOVE THEM RADIATING INFRARED AT THEM, RATHER THAN THE ICY COLDNESS OF SPACE.

 

 

If it were as simple as the radiative heat transfer being equal to the temperature difference to the fourth power, wouldn't the entire OTA glow like an incandescent bulb?  Because the temperature difference between the OTA and space is larger than the temperature difference between the bulb filiment and the interior of a house.

 

NO - RADIATIVE EMISSIONS DONT DEPEND ON THE DIFFERENCE BETWEEN TEMPERATURES - EVERYTHING RADIATES. THE ONLY DIFFERENCE IS HOW MUCH HEAT THEY GET BACK.

 

Here is the way I'm thinking about it:

 

1) Once a photon leaves the surface of the objective, its gone.  The environment will not determine the rate at which photons leave.  It does determine the rate at which photons are ABSORBED by the objective

 

CORRECT,AND CONTRADICTING YOUR SENTENCE ABOVE.

 

2) Shielded or unshielded, radiative processes lead to a net loss of heat energy by the objective, because in both cases the environment is (initially) colder than the objective, and thus less radiative

 

TRUE - BUT DONT FORGET SPACE IS ALWAYS COLDER AND THIS DOMINATES.

 

3) Because the metal tube has a much larger surface area than the objective, it will lose heat more rapidly and cool faster.  Because it is an excellent conductor of heat, conductive heat loss to the tube will likely be larger than direct radiative heat loss from the objective.  So items 1 and 2 are second order effects. 

 

THIS COULD BE TRUE WHERE THE OBJECTIVE/MIRROR IS IN EXCELLENT THERMAL CONTACT WITH THE TUBE. I DOUBT THIS IS OFTEN THE CASE BUT IM A REFLECTOR GUY SO DONT KNOW FOR SURE.

 

If I wrap a 100 degree piece of heating tape around my metal tube in the vicinity of the objective, heat is transferred into the glass at a rate sufficient to keep it from dewing up, irrespective of whether or not it is shielded.  So wouldn't we expect a very rapid loss of heat in the other direction if we don't apply the heating tape and the tube cools to 40 degrees?

 

YEAH WE MIGHT! ITS A GOOD POINT. SO FOR REFRACTORS IT MIGHT MAKE SENSE TO INSULATE THE TUBE.


Edited by happylimpet, 20 February 2020 - 08:58 AM.


#11 Jon Isaacs

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Posted 20 February 2020 - 08:58 AM

If it were as simple as the radiative heat transfer being equal to the temperature difference to the fourth power, wouldn't the entire OTA glow like an incandescent bulb?  Because the temperature difference between the OTA and space is larger than the temperature difference between the bulb filiment and the interior of a house.

 

 

There are emmissivities involved but if i remember correctly it's not the temperature differential to the 4th power, it's the difference between the temperatures to the 4th power:

 

Tscope4 - Tspace4..

 

And yes, at infrared wave lengths, your scope does glow.

 

I'm sure Red can supply better explanations, he's an engineer who specializes in heat transfer analysis. I had those courses almost 40 years ago.

 

Jon



#12 aa5te

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Posted 20 February 2020 - 09:06 AM

Why can't we just say that dew is scared of the the dew shield because the dew shield is packing heat? Huh?


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#13 JPSTAR

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Posted 20 February 2020 - 09:28 AM

If I remember my physics correctly, radiative cooling is related to T4.  This is huge because the night sky is very cold in terms of radiation, something like -40C or colder.

 

The dew shield is only a few degrees cooler than the objective so there's very little radiative heat transfer between the objective and the dew shield.

 

Jon

From my Chem E classes some 60 years ago, I recall using -40C as an effective radiation temperature of the night sky.  Apparently it depends on a lot of factors such as cloud cover and height, surface temperature of the earth, and humidity.  See the posts in the following for a more detailed explanation:

 

https://physics.stac...urface-of-earth


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#14 WadeH237

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Posted 20 February 2020 - 09:41 AM

What actually happens is that for a limited period of time, the shield is able to reduce the amount of humid air passing over the cold surface of the lens.  The water of the condensate has to come from somewhere!

 

Thus heated dew shields are needed in the long run when air temperatures fall to near or below the dew point.

If this were true, then the corrector or lens would never dew up.

 

Dew will only form when the temperature of the surface is colder than the surrounding air.  This can't happen with conductive cooling.  You need radiative cooling.

 

You can read about it here.


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#15 John Tucker

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Posted 20 February 2020 - 09:42 AM

YES IT DOES, BUT IMPORTANTLY THE OBJECTIVE ALSO RECEIVES RADIATED HEAT BACK FROM THE DEW SHIELD. REMEMBER, FREEZING TEMPS FEEL COLD TO US MAMMALS, BUT A FREEZING NIGHT IS HOT COMPARED TO MOST OF THE UNIVERSE.

 

But it takes about 20 Wh of power applied to a heating tape to prevent my objective from dewing up over a 10 hour session.  The dew shield doesn't have that much heat capacity, and if it did most of it would be lost by radiation in other directions (outward).  The workaround is for the shield to extract heat from the air, but it has to do this faster then the large metal tube does and that seems unlikely.

 

 

THERE IS ALWAYS SUFFICIENT CIRCULATION TO PROVIDE PLENTY OF DEW IF SURFACES ARE COLD. IN FACT AIR FLOW CAN KEEP SURFACES DEW FREE (COLD WINDY NIGHTS I NEVER HAVE ANY DEW) AS THEY KEEP SURFACES AT AMBIENT.

 

The effect of wind is to "warm" the tube so that it doesn't cool significantly below the temperature of the surrounding air.  Also the primary on my Newt never dews up, even when the outer surface of the mirror has dew on it. 

 

 

NOPE. LOOK AT OPEN AND CLOSED TUBE NEWTONIANS - THE ULTIMATE DEW SHIELD! PRIMARIES ESSENTIALLY NEVER DEW UP AND NOONE HEATS THEIR TUBES.

 

The outer surfaces do. The inner surfaces don't get enough air circulation to pick up a lot of condensate. 

 

 

NOPE, THIS IS BECAUSE THE STUFF INSIDE IS KEPT WARM BY HAVING A PLASTIC COVER AT 273K ABOVE THEM RADIATING INFRARED AT THEM, RATHER THAN THE ICY COLDNESS OF SPACE.

 

As with the dew shield, for the cover to transfer heat to the items in the box all night long and not fall to liquid nitrogen temperatures itself would require a much larger heat capacity than a 4 oz plastic lid has.  


Edited by John Tucker, 20 February 2020 - 09:55 AM.


#16 TOMDEY

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Posted 20 February 2020 - 10:29 AM



If I remember my physics correctly, radiative cooling is related to T4.  This is huge because the night sky is very cold in terms of radiation, something like -40C or colder.

 

The dew shield is only a few degrees cooler than the objective so there's very little radiative heat transfer between the objective and the dew shield.

 

Jon

Yes! It all comes out in the wash. So it's the delta Tabsloute differential (thermal power loss proportional to Tlens4 - Tsky4 ). The reason that the passive dew shield helps, is because it profoundly reduces the Solid Angle of the sky seen by the radiatively-lambertian lens. Other nuances enter into it, like the emissivity spectra of sky vs lens, etc... but the diminished étendue is the biggest driver in passive dew shield efficacy. That's a function of the shield's Aspect Ratio = H/D, where H is its height and D is its diameter. deriving... deriving... OK, the radiometric loss % Improvement, over no dew shield whatsoever, is simply proportional to 2K/root(1+4K2), where K is the aspect ratio. Table:

 

aspect ratio       % Improvement

  K = H/D              vs no shield

 

   (infinity)                   100%

       3                           99

       2                           97

       1                           89

      0.5                         71

 (no shield)                   0

 

And we see that a shield doesn't have to be very long to help a lot! This also explains why domed observatories are dew-free. Indeed, mine would be completely dripping water, approaching dawn, on the outside --- but still entirely dry on the inside... interior surface of dome and all equipment, including the top of the scope. Ummm... Oh, yeah... those % improvements are benchmarked relative to worst-case heat loss: completely clear sky, from zenith to horizon, scope pointing straight up. In actual practice, trees, buildings, pointing, etc. also help a lot.

 

[I worked radiometry for satellite thermal control stuff, back in my working days and nights... now those things shed heat like crazy!]    Tom


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#17 happylimpet

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Posted 20 February 2020 - 10:40 AM

YES IT DOES, BUT IMPORTANTLY THE OBJECTIVE ALSO RECEIVES RADIATED HEAT BACK FROM THE DEW SHIELD. REMEMBER, FREEZING TEMPS FEEL COLD TO US MAMMALS, BUT A FREEZING NIGHT IS HOT COMPARED TO MOST OF THE UNIVERSE.

 

But it takes about 20 Wh of power applied to a heating tape to prevent my objective from dewing up over a 10 hour session.  The dew shield doesn't have that much heat capacity, and if it did most of it would be lost by radiation in other directions (outward).  The workaround is for the shield to extract heat from the air, but it has to do this faster then the large metal tube does and that seems unlikely.

 

The heat capacity of the heatshield is pretty much irrelevant. Because its immersed in earths atmosphere, it can radiate all it likes but will never get more than a few degrees below ambient. Thats enough for dew to form, but not enough to stop it being a powerful emitter of infrared heat that warms your objective. Crazy, but true.

 

THERE IS ALWAYS SUFFICIENT CIRCULATION TO PROVIDE PLENTY OF DEW IF SURFACES ARE COLD. IN FACT AIR FLOW CAN KEEP SURFACES DEW FREE (COLD WINDY NIGHTS I NEVER HAVE ANY DEW) AS THEY KEEP SURFACES AT AMBIENT.

 

The effect of wind is to "warm" the tube so that it doesn't cool significantly below the temperature of the surrounding air.  Also the primary on my Newt never dews up, even when the outer surface of the mirror has dew on it. 

 

Thats right.

 

NOPE. LOOK AT OPEN AND CLOSED TUBE NEWTONIANS - THE ULTIMATE DEW SHIELD! PRIMARIES ESSENTIALLY NEVER DEW UP AND NOONE HEATS THEIR TUBES.

 

The outer surfaces do. The inner surfaces don't get enough air circulation to pick up a lot of condensate. 

 

The outer surfaces do because of radiative cooling, and unless lined the inner surfaces also dew up. Mine do anyway.

 

NOPE, THIS IS BECAUSE THE STUFF INSIDE IS KEPT WARM BY HAVING A PLASTIC COVER AT 273K ABOVE THEM RADIATING INFRARED AT THEM, RATHER THAN THE ICY COLDNESS OF SPACE.

 

As with the dew shield, for the cover to transfer heat to the items in the box all night long and not fall to liquid nitrogen temperatures itself would require a much larger heat capacity than a 4 oz plastic lid has.  

 

The heat capacity of the lid is again irrelevant, as soon as it radiates heat away it gets more from the air - all that matters is , as at top, that it cant supercool much because of being immersed in air. So it keeps radiating heat downwards and keeping your eyepieces well above the absolute zero that their upper surfaces are trying hard to reach.



#18 TOMDEY

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Posted 20 February 2020 - 11:01 AM

Bottom line... even a fairly short passive dew shield... helps a LOT! All the other minutia is strolling thru the weeds, but otherwise little more than academically-interesting. Operationally, a modest dew shield plus just a tiny blush of active heating... is generally all it takes. An observatory helps a lot; a domed observatory is essentially immune.

 

Many nights, I close up the dome as dawn approaches, nice and dry inside. And out on the deck, dome is dripping profusely, walking the 1000 ft down to the house, the grassy field is saturated with dew. Pretty nice!    Tom

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#19 happylimpet

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Posted 20 February 2020 - 11:05 AM



Yes! It all comes out in the wash. So it's the delta Tabsloute differential (thermal power loss proportional to Tlens4 - Tsky4 ). The reason that the passive dew shield helps, is because it profoundly reduces the Solid Angle of the sky seen by the radiatively-lambertian lens. Other nuances enter into it, like the emissivity spectra of sky vs lens, etc... but the diminished étendue is the biggest driver in passive dew shield efficacy. That's a function of the shield's Aspect Ratio = H/D, where H is its height and D is its diameter. deriving... deriving... OK, the radiometric loss % Improvement, over no dew shield whatsoever, is simply proportional to 2K/root(1+4K2), where K is the aspect ratio. Table:

 

aspect ratio       % Improvement

  K = H/D              vs no shield

 

   (infinity)                   100%

       3                           99

       2                           97

       1                           89

      0.5                         71

 (no shield)                   0

 

And we see that a shield doesn't have to be very long to help a lot! This also explains why domed observatories are dew-free. Indeed, mine would be completely dripping water, approaching dawn, on the outside --- but still entirely dry on the inside... interior surface of dome and all equipment, including the top of the scope. Ummm... Oh, yeah... those % improvements are benchmarked relative to worst-case heat loss: completely clear sky, from zenith to horizon, scope pointing straight up. In actual practice, trees, buildings, pointing, etc. also help a lot.

 

[I worked radiometry for satellite thermal control stuff, back in my working days and nights... now those things shed heat like crazy!]    Tom

Cheers Tom, your aspect ratio calc here is very informative!



#20 John Tucker

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Posted 20 February 2020 - 12:06 PM

Good convo but we're still left making differing unsupported assertions about whether the difference between the inside of the tube and the outside is due to differences in humid air flux vs. differences in radiative cooling.  

 

I think we still differ on the photon count as well.  

 

As I noted above, the photons leave the objective at the same rate irrespective of whether it is pointed at the sky or anything else.  So the net photon flux depends on variation in the number of photons coming in. 

 

Roughly half the photons leaving the dew shield do so from the outer surface of the shield, and cannot impact the lens.  Those leaving the internal surface mostly strike the opposite face of the shield, as this surface is concave and perpendicular to the lens (which still points at the sky).   This seems like it would lead to a situation in which the great majority of the net photon flux from the shield being directed outward.  

 

The physics goes a bit over my head here, but it is very difficult for me to believe that infrared radiation emitted by a 40 degree, 60 square inch piece of plastic suffices to keep the lens 5 to 10 degrees warmer than it would otherwise be, especially given the configuration here.  Or that it would meaningfully compete with conductive heat loss from a steel tube that has many times the surface are of the lens and the shield combined. 

 

Good calculator here. http://hyperphysics..../radfrac.html  

 

For an 100 mm lens I calculate a shield size of about 120 cm2 internal surface, and I get a radiative emission of about 4W.  

 

How do we figure out how much of this is absorbed by the objective lens?  I'd argue at least half goes out the open end.  So do we assume that the majority of the rest is reabsorbed by the shield itself or is that double counting?


Edited by John Tucker, 20 February 2020 - 01:44 PM.


#21 John Tucker

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Posted 20 February 2020 - 12:32 PM

Here we go.  I'm going to sort this out definitively.   https://www.amazon.c...82219884&sr=1-3



#22 desertlens

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Posted 20 February 2020 - 02:09 PM

Excellent! Thanks all for an in depth lesson in radiative cooling.



#23 John Tucker

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Posted 20 February 2020 - 03:24 PM



Yes! It all comes out in the wash. So it's the delta Tabsloute differential (thermal power loss proportional to Tlens4 - Tsky4 ). The reason that the passive dew shield helps, is because it profoundly reduces the Solid Angle of the sky seen by the radiatively-lambertian lens. Other nuances enter into it, like the emissivity spectra of sky vs lens, etc... but the diminished étendue is the biggest driver in passive dew shield efficacy. That's a function of the shield's Aspect Ratio = H/D, where H is its height and D is its diameter. deriving... deriving... OK, the radiometric loss % Improvement, over no dew shield whatsoever, is simply proportional to 2K/root(1+4K2), where K is the aspect ratio. Table:

 

aspect ratio       % Improvement

  K = H/D              vs no shield

 

   (infinity)                   100%

       3                           99

       2                           97

       1                           89

      0.5                         71

 (no shield)                   0

 

And we see that a shield doesn't have to be very long to help a lot! This also explains why domed observatories are dew-free. Indeed, mine would be completely dripping water, approaching dawn, on the outside --- but still entirely dry on the inside... interior surface of dome and all equipment, including the top of the scope. Ummm... Oh, yeah... those % improvements are benchmarked relative to worst-case heat loss: completely clear sky, from zenith to horizon, scope pointing straight up. In actual practice, trees, buildings, pointing, etc. also help a lot.

 

[I worked radiometry for satellite thermal control stuff, back in my working days and nights... now those things shed heat like crazy!]    Tom

How does this relate to the underlying physics?  

 

The equation seems to be based in "blocking" emission of photons from the lens to the sky.  But of course the photons don't "know" that the shield is present when "deciding" whether to leave the surface of the lens or not.  

 

If we follow them after they leave the lens, they either exit the top of the shield tube or are absorbed by the shield itself.  Is the argument then that they heat the tube, which in turn heats the mirror?  

 

Ultimately I think we have to understand the underlying process to know what equations are appropriate.  And I'm still viewing this as a system in which the shield's main action is to emit photons, not to prevent them from being emitted by the mirror.  At least until someone helps me understand why that is incorrect. 



#24 desertlens

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Posted 20 February 2020 - 04:02 PM

My "aha" in this discussion came when I understood how the instrument could be cooler than ambient and ultimately the dew point. I was thinking in terms of thermal equilibrium rather than radiative cooling. Thanks again all. This clears up a longstanding question for me.


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#25 Redbetter

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Posted 20 February 2020 - 04:29 PM

I disagree with most of the above, and suggest an experiment to differentiate the possibilities.

 

A simple, non-heated dew shield cannot prevent cooling of the objective lens over any meaningful period of time (1 h+) for two reasons

 

1. Radiative energy loss is fairly rapid.  If you wrap a dew shield around the objective, it simply radiates energy to the shield (which is radiating less energy back, as it is colder), and then the shield radiates it to the environment

 

2. Much if not most of the temperature drop is conductive.  The rest of your OTA is radiating energy, and so drops in temperature.  Its mass, and thus its heat capacity, is much greater than that of the objective.  The objective loses heat to the cold parts of the scope by conductive heat loss.  The tube than loses the energy gained by radiative energy loss. This is quite efficient especially if the tube is metal.

 

What actually happens is that for a limited period of time, the shield is able to reduce the amount of humid air passing over the cold surface of the lens.  The water of the condensate has to come from somewhere!

 

Thus heated dew shields are needed in the long run when air temperatures fall to near or below the dew point.

 

The experiment: There are adhesive films that attach to the sides of aquarium tanks that change color with temperature.  Take one of these and put it on the objective of an old scope, with and without a dew shield. I think you will find the tempertures are pretty much the same.

 

Similar effects are seen when you put the lid back on a plastic bin of gear out in the field, even if the lid is only placed on loosely.  10 hours later there is no way the contents of the bin are significantly warmer than the environment.  But the amount of moist air that has been in contact with your stuff is insufficent to leave everything soaking wet. 

It is best to start experiments once one has a better grasp of the relevant factors.  There are several errors in the above analysis.  

 

Radiative heat loss is not all that rapid in this temperature range.  The absolute temperature isn't that high and the difference in absolute temperature is still moderate.  The amount of subcooling is the balance point behind natural convection (or possibly forced convection) and the radiative cooling.  If the temp of exposed surfaces was falling close to that of the effective temperature of the upper atmosphere then you would be correct.  Instead it hovers a few degrees below that of the ambient air.   

 

This is easily demonstrated by the fact that dew is only a problem on some nights, rather than every night.  There is always some moisture in the air and a dew point, but surfaces often don't cool enough to reach that dew point.   Also, a breeze is frequently enough to prevent dewing (the breeze shifts the temperature balance toward ambient because it acts as "forced convection.")

 

The temperature drop is not primarily conductive.  If it were then dew would form in a clearly annular fashion from the outside, progressing inward.   That isn't the way it progresses.  Instead it is more general.  There are other problems with the scenario as well, such as the fact that glass has rather poor heat transfer properties so the long radius from the inside to the outside would retard this even more.

 

Your bin experiment misses the most basic difference:  the bin itself acts as a dew shield for the contents.  Things in my black truck box stay dry because they stay warmer than the surface of the bin exposed to the sky.  The surface subcools early on (particularly since it is black) and this tine of year it often ends up covered in dew or frost.  But the temperature inside the bin doesn't fall as low as the top surface.  I have 24hr temp. gauge inside that logs the highs and lows, and it doesn't fall as low as the ambient.

 

Meanwhile, things placed on top of the bin, or on a table without a cover, radiate to the night sky and dew or frost if conditions are right.  That is why I use a moving blanket as a dew shield to cover my SQM meters, notebooks, and atlases on my table.  They get exposed to air whenever I am using them (quite frequently), but stay far dryer than if they were left exposed.

 

Moving air actually reduces subcooling.  

 

 

If it were as simple as the radiative heat transfer being equal to the temperature difference to the fourth power, wouldn't the entire OTA glow like an incandescent bulb?  Because the temperature difference between the OTA and space is larger than the temperature difference between the bulb filiment and the interior of a house.

 

Here is the way I'm thinking about it:

 

1) Once a photon leaves the surface of the objective, its gone.  The environment will not determine the rate at which photons leave.  It does determine the rate at which photons are ABSORBED by the objective

 

2) Shielded or unshielded, radiative processes lead to a net loss of heat energy by the objective, because in both cases the environment is (initially) colder than the objective, and thus less radiative

 

3) Because the metal tube has a much larger surface area than the objective, it will lose heat more rapidly and cool faster.  Because it is an excellent conductor of heat, conductive heat loss to the tube will likely be larger than direct radiative heat loss from the objective.  So items 1 and 2 are second order effects. 

 

If I wrap a 100 degree piece of heating tape around my metal tube in the vicinity of the objective, heat is transferred into the glass at a rate sufficient to keep it from dewing up, irrespective of whether or not it is shielded.  So wouldn't we expect a very rapid loss of heat in the other direction if we don't apply the heating tape and the tube cools to 40 degrees?

The simplest way to state this is that your model is wrong.  You have to start with a representative model if you ever hope to get results that are consistent with the real world. 

 

First, everything is glowing like an incandescent bulb.  But the temperature is low enough that these emissions are infrared.  Things must be considerably warmer before we start to see a dull red glow.  Radiative losses at ambient temps are low enough that we normally can ignore them, except for the small temperature differences that result in dewing, etc.  

 

The system gets complex in a hurry, so it is best to start with the biggest drivers first, then work from there.  Radiant heat loss is competing with convective cooling (typically natural convection when the air is still.)  The radiant components are in all directions, to and from. 

  • There isn't much radiation down to us from the night sky when it is clear--that -40 C value is the typical normal approximation I have seen.  The value is not zero, but it is not much either.
  • The objective is emitting radiation in all directions, including back into any tube assembly, to the dew shield, and to any exposed night sky (or cap if it is capped.)
  • The dew shield and OTA are also emitting radiation back to the objective.  These are far warmer than the night sky; they will only be subcooled a few degrees below ambient.  If you only had to worry about radiating to the dew shield and OTA then the objective would be unlikely to dew, unless the air was already saturated at or very near the dew point.

Your second point completely discounts what a dew shield or cap actually does.  Tom illustrated the blocking effect based on aspect ratio.  This is directly calculated from the geometry, it is a physical model.  The dew shield quickly subcools where it is exposed to the sky, but the amount of subcooling is still limited by natural convection, so it is only a few degrees colder than ambient.  If this was in a vacuum, then the natural convection would not warm the dew shield and its temp would fall closer to that of the night sky.  Then it would lose most of its ability to prevent subcooling...of course in a vacuum it wouldn't matter.

 

Point three gets the relative contributions to the observed heat transfer backwards.  It is an understandable error, but once a person has done some of the calcs and seen how such systems behave, different results are observed/expected. 

  • Conductive transfer tends to be slower than convective, and in this case the conductive would also be dependent on the radiative loss as well, making it an even weaker actor.  
  • The conductive transfer is reduced by the geometry which is effectively that of a fin. In fintube heat exchange design fin efficiency is a major component in discounting the actual heat transfer coefficient. 
  • The conductive transfer is also reduced by the low thermal conductivity of glass combined with the substantial path length. 

 However, that conduction can be enough to offset the radiant losses when strip heaters are used.


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